Cubic vertex-transitive non-cayley graphs of order 8p

Cubic vertex-transitive non-cayley graphs of order 8p Jin-Xin Zhou, Yan-Quan Feng Department of Mathematics, Beijing Jiaotong University, Beijing, China jxzhou@bjtu.edu.cn, yqfeng@bjtu.edu.cn Submitted: Nov 2, 2010; Accepted: Feb 28, 2012; Published: Mar 9, 2012 Mathematics Subject Classifications: 05C25, 20B25 Abstract A graph is vertex-transitive if its automorphism group acts transitively on its vertices. A vertex-transitive graph is a Cayley graph if its automorphism group contains a subgroup acting regularly on its vertices. In this paper, the cubic vertextransitive non-cayley graphs of order 8p are classified for each prime p. It follows from this classification that there are two sporadic and two infinite families of such graphs, of which the sporadic ones have order 56, one infinite family exists for every prime p > 3 and the other family exists if and only if p 1 (mod 4). For each family there is a unique graph for a given order. Keywords: Cayley graphs, vertex-transitive graphs, automorphism groups 1 Introduction For a finite, simple and undirected graph X, we use V (X), E(X), A(X) and Aut(X) to denote its vertex set, edge set, arc set and full automorphism group, respectively. For u, v V (X), u v means that u is adjacent to v and denote by {u, v} the edge incident to u and v in X. A graph X is said to be vertex-transitive, and arc-transitive (or symmetric) if Aut(X) acts transitively on V (X) and A(X), respectively. Given a finite group G and an inverse closed subset S G \ {1}, the Cayley graph Cay(G, S) on G with respect to S is defined to have vertex set G and edge set {{g, sg} g G, s S}. It is well known that a vertex-transitive graph is a Cayley graph if and only if its automorphism group contains a subgroup acting regularly on its vertex set (see, for example, [25, Lemma 4]). There are vertex-transitive graphs which are not Cayley graphs Supported by the National Natural Science Foundation of China (10901015, 11171020), the Fundamental Research Funds for the Central Universities (2011JBM127, 2011JBZ012), and the Subsidy for Outstanding People of Beijing (2011D005022000005). the electronic journal of combinatorics 19 (2012), #P53 1

and the smallest one is the well-known Petersen graph. Such a graph will be called a vertex-transitive non-cayley graph, or a VNC-graph for short. Many publications have been put into investigation of VNC-graphs from different perspectives. For example, in [13], Marušič asked for a determination of the set NC of non-cayley numbers, that is, those numbers for which there exists a VNC-graph of order n, and to settle this question, a lot of VNC-graphs were constructed in [9, 11, 14, 19, 15, 16, 17, 20, 22, 26]. In [6], Feng considered the question to determine the smallest valency for VNC-graphs of a given order and it was solved for the graphs of odd prime power order. By [19, Table 1], the total number of vertex-transitive graphs of order n and the number of VNC-graphs of order n were listed for each n 26. It seems that, for small orders at least, the great majority of vertex-transitive graphs are Cayley graphs. This is true particularly for small valent vertex-transitive graphs (see [21]). This suggests the problem of classifying small valent VNC-graphs. From [3, 12] all VNC-graphs of order 2p are known for each prime p. Recently, in [30] all tetravalent VNC-graphs of order 4p were classified, and in [28, 29, 31], all cubic VNC-graphs of order 2pq were classified, where p and q are primes. In this paper we shall classify all cubic VNC-graphs of order 8p for each prime p. As a result, there are two sporadic and two infinite families of such graphs, of which the sporadic ones have order 56, one infinite family exists for every prime p > 3 and the other family exists if and only if p 1 (mod 4). For each family there is a unique graph for a given order. 2 Preliminaries In this section, we introduce some notations and definitions as well as some preliminary results which will be used later in the paper. For a regular graph X, use d(x) to represent the valency of X, and for any subset B of V (X), the subgraph of X induced by B will be denoted by X[B]. Let X be a connected vertex-transitive graph, and let G Aut(X) be vertex-transitive on X. For a G-invariant partition B of V (X), the quotient graph X B is defined as the graph with vertex set B such that, for any two vertices B, C B, B is adjacent to C if and only if there exist u B and v C which are adjacent in X. Let N be a normal subgroup of G. Then the set B of orbits of N in V (X) is a G-invariant partition of V (X). In this case, the symbol X B will be replaced by X N. For a positive integer n, denote by Z n the cyclic group of order n as well as the ring of integers modulo n, by Z n the multiplicative group of Z n consisting of numbers coprime to n, by D 2n the dihedral group of order 2n, and by C n and K n the cycle and the complete graph of order n, respectively. We call C n a n-cycle. For two groups M and N, N M denotes a semidirect product of N by M. For a subgroup H of a group G, denote by C G (H) the centralizer of H in G and by N G (H) the normalizer of H in G. Then C G (H) is normal in N G (H). Proposition 1. [10, Chapter I, Theorem 4.5] The quotient group N G (H)/C G (H) is isomorphic to a subgroup of the automorphism group Aut(H) of H. the electronic journal of combinatorics 19 (2012), #P53 2

Let G be a permutation group on a set Ω and α Ω. Denote by G α the stabilizer of α in G, that is, the subgroup of G fixing the point α. We say that G is semiregular on Ω if G α = 1 for every α Ω and regular if G is transitive and semiregular. For any g G, g is said to be semiregular if g is semiregular. The following proposition gives a characterization for Cayley graphs in terms of their automorphism groups. Proposition 2. [25, Lemma 4] A graph X is isomorphic to a Cayley graph on a group G if and only if its automorphism group has a subgroup isomorphic to G, acting regularly on the vertex set of X. 3 Double generalized Petersen graphs In [28, 29, 31], the generalized Petersen graphs (see [27]) were used to construct cubic VNC-graphs with special orders. Let n 3 and 1 t < n/2. The generalized Petersen graph P (n, t) (GPG for short) is the graph with vertex set {x i, y i i Z n } and edge set the union of the out edges {{x i, x i+1 } i Z n }, the inner edges {{y i, y i+t } i Z n } and the spokes {{x i, y i } i Z n }. Note that the subgraph of P (n, t) induced by the out edges is an n-cycle. In this section, we modify the generalized Petersen graph construction slightly so that the subgraph induced by the out edges is a union of two n-cycles. Definition 3. Let n 3 and t Z n {0}. The double generalized Petersen graph DP (n, t) (DGPG for short) is defined to have vertex set {x i, y i, u i, v i i Z n } and edge set the union of the out edges {{x i, x i+1 }, {y i, y i+1 } i Z n }, the inner edges {{u i, v i+t }, {v i, u i+t } i Z n } and the spokes {{x i, u i }, {y i, v i } i Z n } (See Fig. 1 for DP(10,2)). x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 u 0 u 1 u 2 u 3 u 4 u 5 u 6 u 7 u 8 u 9 v 0 v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9 y 0 y 1 y 2 y 3 y 4 y 5 y 6 y 7 y 8 y 9 Figure 1: The graph DP (10, 2) Note that the complete classification of the automorphism groups of GPGs has been worked out in [8], and Nedela and Škoviera [23] have determined all Cayley graphs among GPGs. It is natural to consider the problem of determining all vertex-transitive graphs and all VNC-graphs among DGPGs. The complete solution of this problem may be a topic for our future effort. Here, we just give a sufficient condition for a DGPG being vertex-transitive non-cayley. To do this, we introduce some notations. the electronic journal of combinatorics 19 (2012), #P53 3

In the remainder of this section, we always use p to represent a prime congruent to 1 modulo 4. It is easy to see that λ Z p is a solution of the equation x 2 1 (mod p) (1) if and only if λ has order 4 in Z p. Since p 1 (mod 4), Z p has exactly two elements, say λ, p λ, of order 4. So, in Z p, Eq. (1) has exactly two solutions that are λ and p λ. Note that every solution of Eq. (1) in Z 2p is also a solution of Eq. (1) in Z p. This implies that in Z 2p, Eq. (1) has exactly four pairwise different solutions that are λ, 2p λ, p λ and p + λ. Lemma 4. For any λ 1, λ 2 {λ, 2p λ, p λ, p + λ}, we have DP (2p, λ 1 ) = DP (2p, λ 2 ). Proof. By Definition 3, it is easy to see that if either {λ 1, λ 2 } = {λ, 2p λ} or {λ 1, λ 2 } = {p λ, p + λ}, then DP (2p, λ 1 ) = DP (2p, λ 2 ). To complete the proof, it suffices to show DP (2p, λ) = DP (2p, p + λ). Define a map from V (DP (2p, λ)) to V (DP (2p, p + λ)) as following: f : x i x i, y i y i+p, u i u i, v i v i+p, i Z 2p. It is easy to see that f is a bijection. Furthermore, f maps each of the out edges and the spokes of DP (2p, λ) to an edge of DP (2p, p + λ). For the inner edges, {u i, v i+λ } f = {u i, v i+λ+p } E(DP (2p, λ 2 )). Similarly, {v i, u i+λ } f = {v i+p, u i+λ } = {v i+p, u i+p+p+λ } E(DP (2p, λ 2 )). Thus, f is an isomorphism from DP (2p, λ) to DP (2p, p + λ). Theorem 5. Suppose that V NC 1 8p := DP (2p, λ), where λ is a solution of Eq. (1) in Z 2p. Then V NC 1 8p is a connected cubic VNC-graph of order 8p. Proof. Let X = V NC 1 8p and A = Aut(X). By the definition, it is easy to see that X is connected and has order 8p. Since p 1 (mod 4), one has p 5. If p = 5, with the help of MAGMA [1], X is a cubic VNC-graph. In what follows, assume p > 5. Remember that Eq. (1) has exactly four solutions, namely, λ, 2p λ, p λ and p + λ, in Z 2p. By Lemma 4, we may assume that λ is even. One can easily see that the following maps are permutations on the vertex set of X: α : x i x i+1, y i y i+1, u i u i+1, v i v i+1, i Z 2p, β : x i y i, y i x i, u i v i, v i u i, i Z 2p, γ : x 2i+1 v (2i+1)λ, x 2i u (2i)λ, y 2i+1 v (2i+1)λ+p, y 2i u (2i)λ+p, u 2i x (2i)λ, v 2i x (2i)λ+p, u 2i+1 y (2i+1)λ, v 2i+1 y (2i+1)λ+p, i Z 2p. Also, one may easily see that α and β map each edge of X to an edge. So, α, β Aut(X). Since λ is an even solution of Eq. (1), one has p + λ 2 + 1 0 (mod 2p). For each i Z 2p, we have {x 2i, x 2i+1 } γ = {u (2i)λ, v (2i+1)λ }, {y 2i, y 2i+1 } γ = {u (2i)λ+p, v (2i+1)λ+p }, {u 2i, v 2i+λ } γ = {x (2i)λ, x (2i+λ)λ+p } = {x (2i)λ, x (2i)λ 1 }, {v 2i, u 2i+λ } γ = {x (2i)λ+p, x (2i+λ)λ } = {x (2i)λ+p, x (2i)λ+p 1 }, {u 2i+1, v 2i+1+λ } γ = {y (2i+1)λ, y (2i+1+λ)λ+p } = {y (2i+1)λ, y (2i+1)λ 1 }, {v 2i+1, u 2i+1+λ } γ = {y (2i+1)λ+p, y (2i+1+λ)λ } = {y (2i+1)λ+p, y (2i+1)λ+p 1 }, {x 2i, u 2i } γ = {u (2i)λ, x (2i)λ }, {x 2i+1, u 2i+1 } γ = {v (2i+1)λ, y (2i+1)λ }, {y 2i, v 2i } γ = {u (2i)λ+p, x (2i)λ+p }, {y 2i+1, v 2i+1 } γ = {v (2i+1)λ+p, y (2i+1)λ+p }. the electronic journal of combinatorics 19 (2012), #P53 4

This implies that γ Aut(X). Clearly, {x i, y i i Z 2p } and {u i, v i i Z 2p } are two orbits of α, β on V (X), and γ interchanges these two orbits. Hence, α, β, γ is transitive on V (X). We shall show that A = α, β, γ. Since X has valency 3, the vertex-stabilizer A v is a {2, 3}-group. So, A 2 3+i 3 j p for some integers i, j. As p > 5, the group P = α 2 is a Sylow p-subgroup of A. We claim that P is normal in A. By [7, Theorem 5.1 and Corollary 3.8], this is true for the case when X is symmetric. Suppose X is non-symmetric. Then, 3 A v, and so A is a {2, p}-group. By Burnside s {p, q}-theorem [24, 8.5.3], A is solvable. So, we can take a maximal normal 2-subgroup, say N, of A. Then P N/N A/N, namely, P N A. To show P A, it suffices to prove P P N because then P is characteristic in P N and hence it is normal in A. Consider the quotient graph X N of X relative to the orbit set of N, and let K be the kernel of A acting on V (X N ). Then N K and so K = NK v is a 2-group. The maximality of N gives that K = N. So, A/N Aut(X N ). Clearly, X N has valency 2 or 3, namely, d(x N ) = 2 or 3. Let B V (X N ). If either d(x N ) = 3 or d(x N ) = 2 and d(x[b]) = 1, then the stabilizer N v of v V (X) fixes each neighbor of v. By the connectivity of X, N v fixes all vertices of X. It follows that N v = 1, and hence N is semiregular on V (X). Consequently, N 8. Since p 13, by Sylow theorem, one has P P N, as required. Suppose d(x N ) = 2 and d(x[b]) = 0. Let B 0 and B 1 be two orbits adjacent to B. Since X is cubic, one may assume that d(x[b B 0 ]) = 1 and d(x[b B 1 ]) = 2. Since p is odd, it follows that B = 2 or 4. If B = 2 then X[B B 1 ] = C4. However, since the set of vertices of X at distance 2 from x 0 is N 2 (x 0 ) = {x 2, u 1, v λ, v λ, x 2p 2, u 2p 1 } which has cardinality 6, passing through x 0 there is no cycles of less than 5 in X. This implies that X has girth greater than 4 because it is vertex-transitive. A contradiction occurs. If B = 4, then X[B B 1 ] = C 8 since X has girth greater than 4. So, the subgroup N of N fixing B pointwise also fixes B 0 and B 1 pointwise. By the connectivity of X, N fixes all vertices of X, forcing N = 1. It follows that N Aut(X[B B 1 ]) = D 16. Since p > 5 and p 1 (mod 4), by Sylow Theorem, one has P P N, as required. Now we know the claim is true, that is, P A. Consider the quotient graph X P of X relative to the orbit set of P, and let K be the kernel of A acting on V (X P ). From the construction of X, X P = C8 and the subgraph of X induced by any two adjacent orbits of P is either pk 2 or C 2p. This implies that K acts faithfully on each orbit of P, and hence K Aut(C 2p ) = D 4p. Since K fixes each orbit of P, one has K D 2p. Clearly, A/K is not edge-transitive on X P. It follows that A/K = D 8, and hence A 16p. This implies that A = α, β γ = (Z 2p Z 2 ) Z 4. Now we are ready to finish the proof. Suppose that X is a Cayley graph. By Proposition 2, A has a regular subgroup, say G. Clearly, A : G = 2, and so G is maximal in A. Let Q = α p, β, γ. Then Q is a Sylow 2-subgroup of A. So, Q G, and hence A = GQ. From A = G Q Q G we get that Q G = 8, and hence Q/(Q G) = Z 2. It follows that γ 2 Q G. However, since γ 2 fixes x 0, one has γ 2 / G, a contradiction. the electronic journal of combinatorics 19 (2012), #P53 5

4 Graphs associated with lexicographic products Let n be a positive integer. The lexicographic product C n [2K 1 ] is defined as the graph with vertex set {x i, y i i Z n } and edge set {{x i, x i+1 }, {y i, y i+1 }, {x i, y i+1 }, {y i, x i+1 } i Z n }. In this section, we introduce a class of cubic vertex-transitive graphs which can be constructed from the lexicographic product C n [2K 1 ]. Note that these graphs belong to a large family of graphs constructed in [5, Section 3]. Definition 6. For integer n 2, let X(n, 2) be the graph of order 4n and valency 3 with vertex set V 0 V 1... V 2n 2 V 2n 1, where V i = {x 0 i, x 1 i }, and adjacencies x r 2i x r 2i+1(i Z n, r Z 2 ) and x r 2i+1 x s 2i+2(i Z n ; r, s Z 2 ). Note that X(n, 2) is obtained from C n [2K 1 ] by expending each vertex into an edge, in a natural way, so that each of the two blown-up endvertices inherits half of the neighbors of the original vertex. Definition 7. Let EX(n, 2) be the graph obtained from X(n, 2) by blowing up each vertex x r i into two vertices x r,0 i and x r,1 i. The adjacencies are as the following: x r,s 2i x r,t 2i+1 and x r,s 2i+1 xs,r 2i+2, where i Z n and r, s, t Z 2 (see Fig. 2 for EX(5,2)). x 1,1 0 x 1,1 1 x 1,1 2 x 1,1 3 x 1,1 4 x 1,1 5 x 1,1 6 x 1,1 7 x 1,1 8 x 1,1 9 x 1,0 0 x 1,0 1 x 1,0 2 x 1,0 3 x 1,0 4 x 1,0 5 x 1,0 6 x 1,0 7 x 1,0 8 x 1,0 9 x 0,1 0 x 0,1 1 x 0,1 2 x 0,1 3 x 0,1 4 x 0,1 5 x 0,1 6 x 0,1 7 x 0,1 8 x 0,1 9 0 1 2 3 4 5 6 7 8 9 Figure 2: The graph EX(5, 2) Note that EX(n, 2) is vertex-transitive for each n 2 (see [5, Proposition 3.3]). However, EX(n, 2) is not necessarily a Cayley graph. Below, we shall give a sufficient condition for the graph EX(n, 2) to be vertex-transitive non-cayley. To do this, we need a lemma. Lemma 8. If p > 7 is a prime, then Aut(C p [2K 1 ]) has no subgroups of order 8p. Proof. Set A = Aut(C p [2K 1 ]). It is easily known that A = Z p 2 D 2p. Recall that C p [2K 1 ] has vertex set {x i, y i i Z n } and edge set {{x i, x i+1 }, {y i, y i+1 }, {x i, y i+1 }, {y i, x i+1 } i Z p }. Let K be the maximal normal 2-subgroup of A. Then K = k 0 k 1... k p 1, where k i = (x i y i ) for i Z p. Let α = (x 0 x 1... x p 1 )(y 0 y 1... y p 1 ). It is easy to see that α is an automorphism of C p [2K 1 ] of order p. Set P = α. Suppose to the contrary that A has a subgroup, say G, of order 8p. By Sylow Theorem, one may assume that P G. Since p > 7, Sylow Theorem gives P G. Noting that the electronic journal of combinatorics 19 (2012), #P53 6

GK A, one has A : GK = 1 or 2. Consequently, G K is isomorphic to Z 2 2 or Z 3 2 and is normal in G. It follows that G K C A (P ). However, it is easy to see that C A (P ) K = k 0 k 1... k p 1 = Z 2, a contradiction. Theorem 9. Let p > 3 be a prime. Then the graph V NC 2 8p := EX(p, 2) is a connected cubic VNC-graph of order 8p. Proof. Let X = V NC8p 2 = EX(p, 2) and A = Aut(X). If p = 5 or 7 then by MAGMA [1], X is a connected cubic VNC-graph of order 8p. In what follows, assume that p > 7. By [5, Proposition 3.3], X is vertex-transitive. Clearly, for each j Z p, C 0 j = ( 2j, x0,0 2j+1, x0,1 2j, x0,1 2j+1 ) and C1 j = (x 1,1 2j, x1,1 2j+1, x1,0 2j, x1,0 2j+1 ) are two 4-cycles. Set Ϝ = {C i j i Z 2, j Z p }. From the construction of X, it is easy to see that in X passing each vertex there is exactly one 4-cycle, which belongs to Ϝ. Clearly, any two distinct 4-cycles in Ϝ are vertex-disjoint. This implies that = {V (C i j) i Z 2, j Z p } is an A-invariant partition of V (X). Consider the quotient graph X, and let K be the kernel of A acting on. Then X = Cp [2K 1 ], and hence A/K Aut(C p [2K 1 ]) = Z p 2 D 2p. Noting that between any two adjacent vertices of X there is exactly one edge of X, K fixes each vertex of X and hence K = 1. If X is a Cayley graph, then A = A/K has a regular subgroup of order 8p, and hence Aut(C p [2K 1 ]) would contain a subgroup of order 8p. However, this is impossible by Lemma 8. 5 Classification In this section, we classify all connected cubic VNC-graphs of order 8p for each prime p. Throughout this section, the notations FnA, FnB, etc. will refer to the corresponding graphs of order n in the Foster census of all cubic symmetric graphs [2, 4]. The following is the main result of this paper. Theorem 10. A connected cubic graph of order 8p for a prime p is a VNC-graph if and only if it is isomorphic to F56B, F56C, V NC 1 8p or V NC 2 8p. Proof. By [4], F56B and F56C are cubic symmetric graphs. By MAGMA [1], Aut(F56B) and Aut(F56C) have no subgroups of order 56. It follows from Proposition 2 that F56B and F56C are non-cayley graphs. By Theorems 5 and 9, the graphs V NC8p 1 and V NC8p 2 are connected cubic VNC-graphs of order 8p. To complete the proof, we only need to show necessity. Assume that X is a connected cubic VNC-graph of order 8p. By McKay [18, pp.1114] and [21], all connected cubic vertex-transitive graphs of order 16 or 24 are Cayley graphs. If X is symmetric then by [7, Theorem 5.1], X = F40A, F56B or F56C. By MAMGA [1], F40A = V NC8 5. 1 In what follows, assume that p > 3 and X is non-symmetric. Let A = Aut(X). Since X is non-cayley, A has no subgroups acting regularly on V (X) by Proposition 2. For each v V (X), we have A v = 2 m and A = 2 m+3 p for some positive integer m. By Burnside s p a q b -theorem [24, 8.5.3], A is solvable. For notational convenience, in the remainder of the proof, we always use the following notations. the electronic journal of combinatorics 19 (2012), #P53 7

Assumption For each q {2, p}, use M q to denote the maximal normal q-subgroup of A. Let X Mq be the quotient graph of X relative to the orbit set of M q, and let Ker q be the kernel of A acting on V (X Mq ). We first prove the following claims. Claim 1 Suppose M 2 > 1. Then for any orbit B of M 2, we have X[B] is the null graph. Since p > 2, one has B = 8, 4 or 2, and hence p X M2. This implies that d(x M2 ) 2. If X[B] is not a null graph, then it has valency 1. For each v B, one neighbor of v is in B, and the other two are in the two different orbits of M 2 adjacent to B, respectively. Because Ker 2 fixes each orbit of M 2, (Ker 2 ) v fixes each neighbor of v. By the connectivity of X, (Ker 2 ) v fixes all vertices of X, and hence (Ker 2 ) v = 1. This shows that Ker 2 = M 2 is semiregular. Clearly, X M2 must be a cycle of length l = 8p/ B. The vertex-transitivity of A/M 2 on X M2 implies that A/M 2 contains a subgroup, say G/M 2, acting regularly on V (X M2 ). As a result, G is regular on V (X), a contradiction. Claim 2 Suppose M p > 1. Then X Mp = C8, Ker p = M p A v = D2p and A/Ker p = D8. Furthermore, for any two adjacent orbits, say B, B of M p, we have X[B] = pk 1 and X[B B ] = C 2p or pk 2. Since A = 2 m+3 p, M p is a Sylow p-subgroup of A, and X Mp = 8. So, d(x Mp ) = 3 or 2. Suppose d(x Mp ) = 3. Then the stabilizer (Ker p ) v fixes the neighborhood of v in X pointwise because Ker p fixes each orbit of M p setwise. By the connectivity of X, (Ker p ) v fixes each vertex in V (X), forcing (Ker p ) v = 1. Hence, Ker p = M p. By [21], X Mp is a Cayley graph, and furthermore, either X Mp = Q3, the three dimensional hypercube, or Aut(X Mp ) 16. Note that if X Mp = Q3 then Aut(X Mp ) = S 4 Z 2. Since A/M p = 2 m+3 > 8, A/M p is always a Sylow 2-subgroup of Aut(X Mp ). As X Mp is a Cayley graph of order 8, Aut(X Mp ) has a regular subgroup, say G. By a Sylow Theorem, one may assume that G = G/M p A/M p. This forces that G is regular on V (X), a contradiction. Now we know that d(x Mp ) = 2, namely, X Mp = C8. Then A/Ker p Aut(X Mp ) = D 16. Let V (X Mp ) = {B i i Z 8 } with B i B i+1 for each i Z 8. If some B i contains an edge of X, then the connectivity of X Mp implies that d(x[b i ]) = 1. This forces that B i = p is even, a contradiction. Thus, X[B i ] = pk 1 for every i Z 8. Since X is cubic, for any two adjacent orbits B, B of M p, we have X[B B ] = C 2p or pk 2. Without loss of generality, assume that X[B 0 B 7 ] = pk 2 and X[B 0 B 1 ] = C 2p. Then A/Ker p is not edge-transitive on X Mp, and hence A/Ker p = D8. Since p > 3, the subgroup Ker p of Ker p fixing B 0 pointwise also fixes B 1 and B 7 pointwise. The connectivity of X gives Ker p = 1, and consequently, Ker p Aut(B 0 B 1 ) = D 4p. Since Ker p fixes B 0, one has Ker p = Zp or D 2p. Since A > 8p, it follows that Ker p = D2p and hence A = 16p. Since A/Ker p is regular on V (X Mp ), one has A v = (Ker p ) v = Z2 and Ker p = M p A v. Now we are ready to finish the proof. We distinguish two different cases. Case 1 M p > 1. Since A = 2 m+3 p, one has M p = Zp. Let C = C A (M p ). Then M p C and by Proposition 1, A/C Aut(M p ) = Z p 1. By Claim 2, A/Ker p = D8 and Ker p = M p A v = the electronic journal of combinatorics 19 (2012), #P53 8

D 2p. This means that C v = 1, and hence C is semiregular on V (X). So, C = 2p or 4p. If C = 2p, then C/P = Z 2 is in the center of A/P. Since (A/P )/(C/P ) = A/P Z p 1, A/P is abelian. It follows that A/Ker p = (A/P )/(Kerp /P ) is abelian, a contradiction. So, the only possible is C = 4p. Clearly, C has two orbits, say and on V (X), and the action of C on each of these two orbits is regular. It follows that = {u h h C} and = {v h h C} for some fixed u and v, and furthermore, u h1 u h 2 and v h1 v h 2 for any two distinct h 1, h 2 C. Since is an orbit of C, X[ ] has valency 0, 1 or 2. First, suppose d(x[ ]) = 0. Then X is bipartite. Let the neighbors of u be v h 1, v h 2 and v h 3 where h 1, h 2, h 3 C. Note that C is abelian. For any h C, the neighbors of u h are v hh 1, v hh 2 and v hh 3, and furthermore, the neighbors of v h are u hh 1 1, u hh 1 2 and u hh 1 3. Now it is easy to see that the map α defined by v h u h 1, u h v h 1, h C, is an automorphism of X of order 2. Since C A, C, α = C α has order 8p, implying that C, α is regular on V (X), a contradiction. Next, suppose d(x[ ]) = 1. Let Q be a Sylow 2-subgroup of C. As C is abelian and normal in A, Q is characteristic in C, and hence it is normal in A. Clearly, every orbit of Q has cardinality 4 and is contained in or. Let u h be a neighbor of u, where h C. Clearly, {u, u h } h = {u h, u h2 }. Since d(x[ ]) = 1, one has u h2 = u, implying that h is an involution. It follows that each orbit of Q of C consists of two pairs of adjacent vertices. This is impossible by Claim 1 because Q M 2. Now, suppose d(x[ ]) = 2. Since M p C, each orbit of M p is contained in or. By Claim 2, for any two adjacent orbits B, B of M p, X[B] = pk 1 and X[B B ] = pk 2 or C 2p. Since d(x[ ]) = 2, we must have X[ ] = X[ ] = 2C 2p. Let {x 0, x 1 } be an edge of X[ ]. Then there exists a C such that x 1 = x a 0. From Claim 1 we get a is not an involution. Let a have order s and let x i = x ai 0 with i Z s. Then C 1 = (x 0, x 1,..., x s 1, x 0 ) is an s-cycle. Since X[ ] = 2C 2p, one has s = 2p. Suppose C = Z 4p. Let w be adjacent to x 0, and let {w b, w} E(X[ ]) for some b C. Similar to an argument as above, we get that b has order 2p, and (w, w b, w b2,..., w b2p 1, w) is a 2p-cycle. Since C = Z 4p, one has a = b, and so a = b k for some k Z 2p. This implies that for any i Z 2p, x i = x ai 0 w ai = w bik. Consequently, the subgraph induced by {x i, w bi i Z 2p } has valency 3, contrary to the connectivity of X. Now we know that C = Z 2p Z 2, and hence there is an involution c C \ a. Let y i = x c i with i Z 2p. Since C is abelian, C 2 = (y 0, y 1,..., y s 1, y 0 ) is also a 2pcycle. Clearly, C 1 and C 2 are vertex-disjoint, so X[ ] = C 1 C 2. Note that the edges with one endpoint in and the other endpoint in are independent. Assume that = {u i, v i i Z 2p } so that u i x i and v i y i for i Z 2p. Since X[ ] = 2C 2p, we may assume that u 0 u λ or u 0 v λ for some λ Z 2p {0}. If u 0 u λ, then the subgraph induced by {x i, u i i Z 2p } has valency 3, contrary to the connectivity of X. Thus, u 0 v λ. Since x c i = y i, one has {x i, u i } c = {y i, v i }, and hence u c i = v i. Since c is an involution, one has {u 0, v λ } c = {v 0, u λ }. By Definition 3, X = DP (2p, λ). It is easy to see that C A u is the kernel of A acting on {, }, and A/(C A u ) = Z 2. Let β A be a 2-element interchanging and. Then β 2 C A u. If β 2 C then C, β is regular on V (X), a contradiction. Thus, β 2 = gd where g C and A v = d. the electronic journal of combinatorics 19 (2012), #P53 9

Recalling Ker = P A v = D2p, one has β 2 a 2 β 2 = a 2. It follows that β 1 a 2 β = a 2t for some t Z p satisfying t 2 1 (mod p). Without loss of generality, assume x β 0 = u i for some i Z 2p. Then x β 2 = (x 0 ) a2β = u β 1 a 2 β i = u a2t i = u i+2t. Since the distance between x 0 and x 2 is 2, one has u i+2t = u i+2λ or u i 2λ. It follows that 2t ±2λ (mod 2p), and hence λ ±t (mod p). This shows that λ Z 2p is a solution of Eq. (1). By Lemma 4 and Theorem 5, we have X = V NC8p. 1 Case 2 M p = 1 By the solvability of A, we have M 2 > 1. Let P be a Sylow p-subgroup of A. Then P A but P M 2 /M 2 A/M 2, namely, P M 2 A. If P P M 2, then P is characteristic in P M 2, and hence it is normal in A, a contradiction. Thus, P is not normal in P M 2. Let B be an orbit of M 2. Since p > 2, one has B = 8, 4 or 2, and hence p X M2. This implies that X M2 has valency greater than 1. If d(x M2 ) = 3, then B = 2 or 4, and it is easily seen that M 2 is semiregular, and so M 2 = B. Since p > 3, Sylow Theorem implies that P P M 2, a contradiction. Thus, d(x M2 ) = 2. Also, since A/Ker 2 is transitive on V (X M2 ), Ker 2 is a 2-group. The maximality of M 2 gives Ker 2 = M 2. If B = 8, then X M2 = Cp. By Claim 2, X[B] is a null graph. So, the subgraph induced by any two adjacent orbits is of valency 1 or 2. This forces that X M2 = p is even, a contradiction. If B = 2, then X M2 = C4p, and hence A/M 2 Aut(X M2 ) = D 8p. Since A/M 2 is transitive on V (X M2 ), A/M 2 = D4p, Z 4p or D 8p. This implies that A/M 2 always has a normal subgroup of order 2, contrary to the maximality of M 2. It now only remains to deal with the case when B = 4. In this case, X M2 = C2p and by Claim 2, X[B] = 4K 1. Let V (X M2 ) = {B i i Z 2p } with B i B i+1. Since X is cubic, one may assume that X[B 0 B 1 ] = C 8 or 2C 4 and X[B 0 B 2p 1 ] = 4K 2. Suppose X[B 0 B 1 ] = C 8. The subgroup M2 of M 2 fixing B 0 pointwise also fixes B 1 and B 2p 1 pointwise. The connectivity of X and the transitivity of A/M 2 on V (X M2 ) imply that M2 = 1, and consequently, M 2 Aut(X[B 0 B 1 ]) = D 16. Hence, Aut(M 2 ) is a {2, 3}-group. By Proposition 1, P M 2 /C P M2 (M 2 ) Aut(M 2 ). Since p 5, one has P C P M2 (M 2 ), forcing P P M 2, a contradiction. We now know that X[B 0 B 1 ] is a union of two 4-cycles, say ( 0, 1, x 0,1 0, x 0,1 1 ) and (x 1,1 0, x 1,1 1, x 1,0 0, x 1,0 1 ), where B i = { i, x 0,1 i, x 1,0 i, x 1,1 i } with i = 0 or 1. Remember that X N = (B 0, B 1,..., B 2p 1 ) is a 2p-cycle. Hence, A has an element, say σ, of order p such that Bi σ = B i+2 for each i Z 2p. Without loss of generality, assume σ = (r,s) Z 2 Z 2 (x r,s 0 x r,s 2... x r,s 2 )(x r,s 1 x r,s 3... x r,s 2p 1). Then for each i Z 2p, B i = { i, x 0,1 i, x 1,0 i, x 1,1 i }, and ( 2j, x0,0 2j+1, x0,1 2j, x0,1 2j+1 ) and (x1,1 2j, x 1,1 2j+1, x1,0 2j, x1,0 2j+1 ) are the two 4-cycles of X[B 2j B 2j+1 ] for each j Z p. Note that σ is an automorphism of X. Once the edges between B 2j+1 and B 2j+2 are given, the graph X will be determined. Let u, v be the neighbors of 2i+1 and x0,1 2i+1 in B 2j+2, respectively. If u, v are in the same 4-cycle of X[B 2j+2 B 2j+3 ], then by the connectivity of X, we the electronic journal of combinatorics 19 (2012), #P53 10

2j 2j+1 2j+2 2j+3 x 0,1 2j x 0,1 2j+1 x 0,1 x 0,1 2j+2 2j+3 x 1,0 2j x 1,0 2j+1 x 1,0 x 1,0 2j+2 2j+3 x 1,1 2j x 1,1 2j+1 x 1,1 x 1,1 2j+2 2j+3 2j 2j+1 2j+2 2j+3 x 0,1 2j x 0,1 2j+1 x 0,1 x 0,1 2j+2 2j+3 x 1,0 2j x 1,0 2j+1 x 1,0 x 1,0 2j+2 2j+3 x 1,1 2j x 1,1 2j+1 x 1,1 x 1,1 2j+2 2j+3 Figure 3: Two possible cases get {u, v} = {x 1,0 2i+2, x1,1 2i+2 }. This gives rise to four graphs X i(0 i 4) such that E(X 0 ) = E(X 1 ) = E(X 2 ) = E(X 3 ) = {{x r,s 2i, xr,t 2i+1 {{x r,s 2i, xr,t 2i+1 {{x r,s 2i, xr,t 2i+1 {{x r,s 2i, xr,t 2i+1 }, {xr,s 2i+1, xr+1,s }, {xr,s 2i+1, xr+1,s+1 }, {x0,s 2i+1, x1,s+1 2i+2 }, {x0,s 2i+1, x1,s 2i+2 2i+2 } i Z 2p, r, s, t Z 2 }; 2i+2 } i Z 2p, r, s, t Z 2 }; }, {x1,s 2i+1, x0,s }, {x1,s 2i+1, x0,s+1 2i+2 } i Z 2p, r, s, t Z 2 }; 2i+2 } i Z 2p, r, s, t Z 2 }. Let δ = i Z 2p ( 2i+2, x0,1 2i+2 )(x1,0 2i+2, x1,1 2i+2 ) and γ = i Z 2p ( 2i+2, x0,1 2i+2 ). It is easy to see that δ is an isomorphism from X k to X k+1 with k = 0, 2, and γ is an isomorphism from X 0 to X 3. So, we may assume X = X 0. In this case, X[B 2j B 2j+1 B 2j+2 B 2j+3 ] is the first graph in Fig. 3. Since p > 3, it is easy to check that passing through each vertex of X there is one and only one 4-cycle. Set Ω = {{ i, x 0,1 i }, {x 1,0 i, x 1,1 i } i Z 2p }. Take an arbitrary Ω. Without loss of generality, let = { i, x 0,1 i } for some i Z 4p. For any g A, g B g i = B j for some j Z 4p. Since there is a 4-cycle in X passing through ( i ) g and (x 0,1 i ) g, one has g = { j, x 0,1 j } or {x 1,0 j, x 1,1 j }. It follows that g Ω. Clearly, any two distinct subsets in Ω are disjoint. Then Ω is an A-invariant partition of V (X). From the structure of X we obtain that X Ω = C4p and X[ ] = 2K 1 for each Ω. For notational convenience, let V (X Ω ) = { 0, 1,..., 4p 1 } such that i Ω and i i+1 for each i Z 4p. Since X has valency 3, assume that X[ 0 1 ] = C 4 and X[ 4p 1 0 ] = 2K 2. By the transitivity of A on V (X), X[ 2j 2j+1 ] = C 4 and X[ 2j 1 2j ] = 2K 2 for each j Z 2p. Let i = {x i, y i } for each i Z 4p. From the above analysis we may assume that x i x i+1, y i y i+1, x 2i y 2i+1 and y 2i x 2i+1 for each i Z 4p. Let α : x i x i+2, y i y i+2 (i Z 4p ), β : x i y i, y i x i (i Z 4p ), and γ : x i x 4p+1 i, y i y 4p+1 i (i Z 4p ) be the three permutations on V (X). It is easy to check that α, β and γ are automorphisms of X. Furthermore, α, β, γ = D 4p Z 2 is regular on V (X), a contradiction. Now suppose that u, v are in different 4-cycles of X[B 2j+2 B 2j+3 ]. By [5, Proposition 3.1], we may assume that X[B 2j B 2j+1 B 2j+2 B 2j+3 ] is the second graph in Fig. 3 In this case, E(X) = {{x r,s 2i, xr,t 2i+1 }, {xr,s 2i+1, xs,r 2i+2 } i Z 2p, r, s, t Z 2 }. From Definition 7 and Theorem 9, we know that X = V NC 2 8p. the electronic journal of combinatorics 19 (2012), #P53 11

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